Question: Solve the exponential equation for $x$. $3^{7-2x}=\left(\dfrac{1}{27}\right)^{-8}$ $x=$
Solution: The strategy We want to rewrite one of the exponential terms in the equation so that the bases of the two terms are the same. Then, we will be able to equate the exponents and solve for $x$. [Why can we do this?] Matching the bases Let's rewrite ( 1 27 ) − 8 \left(\dfrac{1}{27}\right)\^{ -8} so its base is $3$. ( 1 27 ) − 8 = ( 3 − 3 ) − 8 = 3 − 3 ⋅ − 8 = 3 24 Rewrite 1 27 as 3 − 3 Since ( a n ) m = a n ⋅ m \begin{aligned} \left(\dfrac{1}{27}\right)\^{ -8} &= (3^{-3})\^{ -8}&&&&\text{Rewrite } \dfrac{1}{27} \text{ as }3^{-3} \\\\ &=3\^{ -3\ \cdot\ -8} &&&&\text{Since }(a^n)^m=a^{n\cdot m}\\\\ &=3\^{ 24} \end{aligned} [Can we choose another base?] Solving the equation We obtain the following equation. 3 7 − 2 x = 3 24 3\^{7-2x}=3\^{ 24} Now we can equate the exponents and solve for $x$. $\begin{aligned} 7-2x&=24\\\\\\ x &=-\dfrac{17}{2}\end{aligned}$ The answer The answer is $x=-\dfrac{17}{2}$. You can check this answer by substituting $\it{x=-\dfrac{17}{2}}$ in the original equation and evaluating both sides.